How would we evaluate the definite integral $\int_0^{e^{\frac{1}{e}}}\frac {-W(-\ln(x))}{\ln(x)}dx$

How would we evaluate the definite integral: $$\int_0^{e^{\frac{1}{e}}}\frac {-W(-\ln(x))}{\ln(x)}dx$$

Here $$W$$ is the Lambert W function. More information about this function can be found here.

Edit 1: Found a mistake in the limits so corrected it.

Edit 2: Here are some alternate forms that I derived that might help:

$$\int_{-1/e}^{\infty} e^{(-u-W(u))}du$$ $$-\int_{-1/e}^{\infty} \frac{W(x)}{Inv(W(x))}du$$ $$\int_{-1/e}^{\infty} \frac { \sum_{n=0}^{ \infty } \frac {(-1)^{n-1}n^{n-1}x^{n}} {n!}} {x\sum_{n=0}^{ \infty }\frac{x^n}{n!} } dx$$

Edit 3: After a bit of more investigation, I have managed to further simplify the integral: $$\int_0^{e^{\frac{1}{e}}}\frac {-W(-\ln(x))}{\ln(x)}dx = -\int_{-1/e}^{\infty} \frac{W(u)}{u(e^u)}du = -\int_{-1/e}^{\infty} \sum_{n=0}^{\infty} \frac {(-1)^nT(n)u^n }{n!}du = \sum_{n=0}^{\infty}\left[ \frac {(-1)^nT(n)u^{(n+1)} }{n!(n+1)} \right]_{-1/e}^{\infty}$$

where T(n) counts the number of forests of rooted labeled trees using labels in a subset of {1,⋯,n} , which is also the hyperbinomial transform of the constant sequence of 1's. (See A088957.)

Now the problem is that I don't know how to evaluate this summation due to T(n), and even wolfram alpha cant do it. Any help on evaluating this summation?

• The proper spelling is definite, not definate. I fixed it for you. – Cameron Williams Mar 28 at 11:32
• ahh a typo thank you – Rithik Kapoor Mar 28 at 11:38
• Since the function $W$ can't be expressed in terms of elementary functions, you could try evaluating the integral numerically. – unseen_rider Mar 28 at 12:03
• An equivalent expression is $$\int_{-\frac1{e}}^\infty\exp(-u-W(u))\mathrm du$$ which is effectively a Laplace transform plus a dangling constant. – J. M. is a poor mathematician Mar 28 at 13:38
• @J.M.isnotamathematician Yes I just derived that but have no idea how to proceed forward. – Rithik Kapoor Mar 28 at 13:43

If we consider the function $$f(x)=-\frac {W(-\ln(x))}{\ln(x)}$$ we can notice that $$xf(x)$$ looks like a power law.

Then, I thought that we could get some reasonable approximations writing $$xf(x)=\sum_{n=1}^p a_n x^{bn}$$ Using $$p=4$$, the results from a quick and dirty regression leads to $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a_1 & +0.49967 & 0.00119 & \{+0.49731,+0.50203\} \\ a_2 & +0.58096 & 0.00575 & \{+0.56954,+0.59239\} \\ a_3 & +1.27106 & 0.04029 & \{+1.19105,+1.35106\} \\ a_4 & -0.72606 & 0.02198 & \{-0.76971,-0.68242\} \\ b & +1.13077 & 0.00065 & \{+1.12948,+1.13206\} \end{array}$$ and computing $$I(t)=-\int_0^t \frac {W(-\ln(x))}{\ln(x)}\,dx$$ we should have the following results (not fantastic, I agree !) $$\left( \begin{array}{ccc} t & \text{approximation} & \text{exact} \\ 0.05 & 0.015218 & 0.015313 \\ 0.10 & 0.034027 & 0.034118 \\ 0.15 & 0.054945 & 0.055038 \\ 0.20 & 0.077617 & 0.077711 \\ 0.25 & 0.101866 & 0.101959 \\ 0.30 & 0.127595 & 0.127687 \\ 0.35 & 0.154749 & 0.154842 \end{array} \right)$$

For sure, this could be made better at the price of more parameters using for example $$xf(x)=\sum_{n=1}^p a_n x^{b_n}$$

• Thanks for this man. This could help. Also I recently edited the question in which I found the indefinite integral in summation form. Do you have any idea on how to evaluate this summation? – Rithik Kapoor Mar 29 at 7:07
• @RithikKapoor. No idea at all ! – Claude Leibovici Mar 29 at 7:10
• Also I am not familiar with the T(n) function and only used it as I found this relation in a math paper. Do you know whether it is the same as tree(n) function. If so then wouldn't it produce extremely big numbers when n>2? – Rithik Kapoor Mar 29 at 7:15

Change variables $$y=-W(-\log x)$$ to convert it to the integral $$\int_{-\infty}^{1}(1-y)\exp(ye^{-y})\;dy$$ But that is still not elementary of course.